已知x²-4x+1=0,求x&sup4+1/x&sup4的值

问题描述:

已知x&sup2-4x+1=0,求x&sup4+1/x&sup4的值

x²-4x+1=0,则x≠0所以方程两边可以同时除以x,得x-4+1/x=0则x+1/x=4x&sup4+1/x&sup4=(x²+1/x²)²-2=[(x+1/x)²-2]²-2=(4²-2)²-2=14²-2=194